'Speedy Summations' printed from http://nrich.maths.org/

Amrit explained how to find a formula for the sum of consecutive numbers, and then adapted it to answer the other questions:

Pair the values as follows:

1

2

3

...

n

n

n-1

n-2

...

1

Each of the columns sums to $n + 1$, and there are $n$ columns, so the sum
should be $n(n + 1)$. However, as each column is counted twice (once in each
possible order of the two values), the sum is $\frac{n(n + 1)}{2}$.

Can you find the sum of all the integers less than 1000 which are not divisible by 2 or 3?

We have to consider a way to express the numbers which aren't divisible by two or three.

The first few are: $1, 5, 7, 11, 13, 17, 19, 23, 25...$
We can see that we have repeating differences of $4$ and $2$, but if we
separate them up we can get two sequences with differences of $6$:

$, 7, 13, 19, 25...$ and $5, 11, 17, 23...$

For the first sequence, $a = 1, d = 6$ and the $n^{th}$ term is the biggest number
of the form $6m + 1$ that is still less than $1000$. I found this number to be
$997$.

$1 + 6(n-1) = 997$
$n - 1 = 166$
$n = 167$

So $S_167 = 83.5(2 + 996) = 83333$

For the second sequence I did the same thing, but with $a = 5$ and the $n^{th}$
term being the largest number of the form $6m + 5$ but still less than $1000$,
which is $995$.

$5 + 6(n-1) = 995$
$(n-1) = 165$
$n = 166$

So $S_166 = 83(10 + 990) = 83000$

Therefore the sum of all integers that are not divisible by $2$ or $3$ and are
less than $1000$ is $83000 + 83333 = 166,333$

Can you find a set of consecutive positive integers whose sum is 32?

Consecutive => d = 1

So $\frac{n}{2}(2a + n - 1) = 32$, so $n(2a + n - 1) = 64$

Since n > 0, I decided to work with increasingly large values of n up
until the point where they began to return smaller and smaller decimals or
negatives or any other sign that I wouldn't get any more valid values of
a.

$n = 1$
$2a = 64$
$a = 32$ <- this is only one number though so it doesn't really count!

As n -> infinity, the constant term in the numerator and
the n term in the numerator become arbitrarily small compared to the -n^2.
So a therefore approaches (-n^2)/2n = -n/2 and so a will continue getting
more and more negative and so we can stop checking n terms and conclude
that there are no sets of consecutive positive integers whose sum is 32.

Rajeev had a different approach for the penultimate question:

Can you find the sum of all the integers less than 1000 which are not divisible by 2 or 3?

Here we first find the sum of all integers which are divisible by 2 and then find the sum of all integers which are divisible by 3 and add these two up and then subtract the sum of the integers that are divisible by 6

$999 \times 1000 /2 = 499500$

Multiples of 2:
$499 \times 1000 /2 = 249500$

Multiples of 3:
$333 \times 1002/2 = 166833$

As multiples of 2 and 3 have duplications they must be removed by counting the number of multiples of 6:
$166 x 1002/2 = 83166$

$499500 – (249500 +166833 – 83166)= 166333$
The sum of all the integers less than 1000 which are not divisible by 2 or 3 is166333

Well done to Abi from Wilberforce Sixth Form College who sent in a clear solution similar to Josh's and Amrit's.